Index Set Proof Let $I$ be a nonempty index set and let $\mathcal{F}$ be a set of subsets $T_\alpha$ of a universal set $U$ indexed by $I$.
    That is, let $\mathcal{F} = \{T_\alpha \mid \alpha\in I\}$.
    Let $X\subseteq U$. 
Prove for each $\beta\in I$, $\bigcap\limits_{\alpha \in I} T_\alpha \subseteq T_\beta$
Proof.  Let $x \in I$. This means $x \in \beta$. Therefore, $x\in$ the set of subsets $T_\alpha$ and $T_\beta$. Since $x\in$ $T_\alpha$ and $T_\beta$, then $T_\alpha \subseteq T_\beta$
Am I on the right track or falling off the edge? Should I use element chasing?
Now, how could I prove $X\cap \left( \bigcup\limits_{\alpha\in I} T_\alpha \right) = \bigcup\limits_{\alpha\in I} (X\cap T_\alpha)$ under the same circumstances?
First, I would let $y \in X\cap \left( \bigcup\limits_{\alpha\in I} T_\alpha \right)$. Then $y\in x$ and $y \in \left( \bigcup\limits_{\alpha\in I} T_\alpha \right)$. Then $y \in T_\beta$ for some $\beta \in I$. $y\in x \cap T_\beta$
Have I started it right and where do i go from here?
 A: PROOF: Let $\beta \in I$. Let $x \in \bigcap_{\alpha \in I} T_{\alpha}$. By definition, this means that, for every $\alpha \in I$, $x \in T_{\alpha}$. In particular, $\beta \in I$, so $x \in T_{\beta}$. Since $x$ was arbitrary, we have proved that $\bigcap_{\alpha \in I} T_{\alpha} \subseteq T_{\beta}$. And since $\beta$ was arbitrary, we have proved the claim. QED
Remember: the structure of your proof is based on the logical form of the statement you wish to prove. The logical form here is:
$$
\forall \beta \in I \, \biggl( \bigcap_{\alpha \in I} T_{\alpha} \subseteq T_{\beta} \biggr),
\quad \text{which is equivalent to} \quad
\forall \beta \in I \, \forall x \biggl( x \in \bigcap_{\alpha \in I} T_{\alpha} \;\to\; x \in T_{\beta} \biggr). 
$$
Based on the latter, the structure of your proof should look like so:

"Let $\beta \in I$."
        "Let $x$ be arbitrary."
              "Suppose $x \in \bigcap_{\alpha \in I} T_{\alpha}$."
              [Proof that $x \in T_{\beta}$ goes here.]
        "Since $x$ be arbitrary, we have proved that $\bigcap_{\alpha \in I} T_{\alpha} \subseteq T_{\beta}$."
  "Since $\beta$ was arbitrary, we have proved that $\forall \beta \in I \, \bigl( \bigcap_{\alpha \in I} T_{\alpha} \subseteq T_{\beta} \bigr)$."

And that, my friend, is exactly how my proof is written.
